Question

Calculate the \(\mathrm{pH}\) of a \(0.42 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) solution.

Short answer

The \(\mathrm{pH}\) of the \(0.42 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) solution is approximately 0.38

Step-by-step solution

Determine the dissociation of \(\mathrm{NH}_{4} \mathrm{Cl}\)

When \(\mathrm{NH}_{4} \mathrm{Cl}\) dissolves in water, it dissociates into \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{Cl}^{-}\). The \(\mathrm{NH}_{4}^{+}\) ions will behave as an acid and react with water to form \(\mathrm{NH}_{3}\) and \( H_{3}O^{+}\). Write the equation for this reaction as: \(\mathrm{NH}_{4}^{+} + H_{2}O \rightarrow \mathrm{NH}_{3} + H_{3}O^{+}\)

Calculate the concentration of \(\mathrm{NH}_{4}^{+}\) and \( H_{3}O^{+}\)

The concentration of \(\mathrm{NH}_{4}^{+}\) is equal to the initial molarity of the \(\mathrm{NH}_{4} \mathrm{Cl}\), 0.42 M. In the reaction, one mole of \(\mathrm{NH}_{4}^{+}\) produces one mole of \( H_{3}O^{+}\), so the concentration of \( H_{3}O^{+}\) is also 0.42 M.

Calculate the \(\mathrm{pH}\)

The \(\mathrm{pH}\) of the solution can be calculated using the \(\mathrm{pH}\) formula: \(\mathrm{pH} = -\log[H^{+}]\). Substitute the concentration of \( H_{3}O^{+}\) into the \(\mathrm{pH}\) formula to calculate the \(\mathrm{pH}\) of the solution: \(\mathrm{pH} = -\log(0.42)\).